Blank of water is a physical change because boiling the water is blank from a liquid to a gas

5.Are pure solids included in equilibrium expressions? Explain your answer.

tatiyna

Answer:

Pure solids or liquids are excluded from the equilibrium expression because their effective concentrations stay constant throughout the reaction. The concentration of a pure liquid or solid equals its density divided by its molar mass.

Explanation:

Hope this helps!!!

8 0

3 years ago

Which are forms of frozen water? Check all that apply. dew frost hail rain sleet

MariettaO [177]

When moist air cools below its dew point on a cold surface it forms dew. It is in the liquid state.

Rain is again a form of precipitation in which water is in the liquid state.

When air temperature is below freezing, the precipitation that results is referred to as frost. Hail is a form of precipitation in the ice balls whereas as sleet is a mixture of rain and snow.

Ans: Forms of frozen water- frost, hail and sleet

7 0

3 years ago

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why do you think the purple cabbage indicator used in this experiment is called "red cabbage"? where is it typically used in foo

tatiyna

Answer:

Red cabbage juice contains a natural pH indicator that changes colors depending on the acidity of the solution. The pigment in red cabbage that causes the red color change is called flavin (an anthocyanin). ... At a lower pH, more hydrogen ions are in solution, and therefore the solution is acidic

Actually I don't know but I tried so I hope it helped u!

4 0

3 years ago

Determine if each of the following statements about chiral molecules is True or False and select the best answer from the dropdo

oee [108]

Answer:

See explanation

Explanation:

-) True or False: All R stereocenters are dextrorotatory.

The absolute configuration is based is in specific rules that are not related to the ability to deflect polarized light. FALSE

-) True or False: Chiral centers in organic molecules will have 4 different groups attached to a carbon.

A chiral carbon by definition is a carbon with 4 groups. TRUE

-) True or False: A racemic mixture has an optical activity of 0.

In a racemic mixture, we have equal amounts of enantiomers, and this cancels out the optical activity. TRUE

-) True or False: Normal linear amines can be chiral centers.

In primary amines, we have 2 hydrogens. Therefore all the groups can not be different. So, is TRUE

-)True or False: Compound C has an optical activity of 0.

We need to know the structure of the compound

-)True or False: If the R isomer of a molecule has an optical rotation of + 25.7 degree then the S isomer of the molecule will have an optical rotation of -25.7 degree.

If we have the exact opposite (as we have in this case) the magnitud of the optical activity value will remain and the sign will change. TRUE

-)True or False: A CN is a higher priority than a CH2OH.

In this case, a carbon is directly bonded to the chiral carbon. The carbon on CN is bonded to a nitrogen atom and the carbon on CH2OH is bonded to an oxygen. So, the CH2OH will have more priority because O has a higher atomic number. FALSE

-)True or False: All molecules with chiral centers are optically active.

We can have for example mesocompounds in which the optical activity is canceled out due to symmetry planes. FALSE

-)True or False: To have an enantiomer a molecule must have at least two chiral centers.

A pair of enantiomers is made between at least 1 chiral carbon. Enantiomer R and enantiomer S. FALSE

-)True or False: Chiral molecules are always optically active.

We can have racemic mixtures or mesocompounds with chiral carbons but without optical activity. FALSE

-)True or False: A CH2CH2Br is higher priority than a CH2F.

The "Br" atom is bonded in the third carbon (respect to the chiral carbon) and the "F" atom is bonded to the second carbon. Therefore CH2F has more priority than CH2CH2Br. FALSE

-)True or False: Meso molecules with two stereocenters have a R,S configuration.

On the compounds with R and S configuration at the same time can have symmetry planes so, we will not have optical activity. TRUE

True or False: Diastereomers have the same physical properties except in a chiral environment.

All diastereomers have the same physical properties. TRUE

True or False: Compound H has an optical activity of 0.

We have to have the structure of the compound.

True or False: A C=C double bond is higher priority than a -CH(CH3)2.

In the case of C=C, we can say that is equivalent to two carbon bonds without hydrogens. Therefore C=C has higher priority. TRUE

6 0

3 years ago

HELP WITH CHEMISTRY PLEASE!

maria [59]

Answer:

1) 1.52 atm.

2) 647.85 K.

3) 20.56 L.

4) 1.513 mole.

5) 254.22 K = -18.77 °C.

Explanation:

In all this points, we should use the law of ideal gas to solve this problem: PV = nRT.
Where, P is the pressure (atm), V is the volume (L), n is the number of moles, R is the general gas constant (0.082 L.atm/mol.K), and T is the temperature (K).

1) In this point; n, R, and T are constants and the variables are P and V.

P and V are inversely proportional to each other that if we have two cases we get: P1V1 = P2V2.

In our problem:

P1 = ??? (is needed to be calculated) and V1 = 45.0 L.

P2 = 5.7 atm and V2 = 12.0 L.

Then, the original pressure (P1) = P2V2 / V1 = (5.7 atm x 12.0 L) / (45.0 L) = 1.52 atm.

2) In this case, n and R are the constants and the variables are P, V, and T.

P and V are inversely proportional to each other and both of them are directly proportional to the temperature of the gas that if we have two cases we get: P1V1T2 = P2V2T1.

In our problem:

P1 = 212.0 kPa, V1 = 32.0 L, and T1 = 20.0 °C = (20 °C + 273) = 293 K.

P2 = 300.0 kPa, V2= 50.0 L, and T2 = ??? (is needed to be calculated)

Then, the temperature in the second case (T2) = P2V2T1 / P1V1 = (300.0 kPa x 50.0 L x 293 K) / (212.0 kPa x 32.0 L) = 647.85 K.

3) In this case, P, n and R are the constants and the variables are V, and T.

V and T are directly proportional to each other that if we have two cases we get: V1T2 = V2T1.

In our problem:

V1 = 25.0 L and T1 = 65.0 °C + 273 = 338 K.

V2 = ??? (is needed to be calculated) and T2 = 5.0 °C + 273 = 278 K.

Herein, there is no necessary to convert T into K.

Then, the volume in the second case (V2) = V1T2 / T1 = (25.0 L x 278 °C) / (338 °C) = 20.56 L.

4) We can get the number of moles that will fill the container from: n = PV/RT.

P = 250.0 kPa, we must convert the unit from kPa to atm; 101.325 kPa = 1.0 atm, then P = (1.0 atm x 250.0 kPa) / (101.325 kPa) = 2.467 atm.

V = 16.0 L.

R = 0.082 L.atm/mol.K.

T = 45 °C + 273 = 318 K.

Now, n = PV/RT = (2.467 atm x 16.0 L) / (0.082 L.atm/mol.K x 318 K) = 1.513 mole.

5) In this case, V, n and R are the constants and the variables are P, and T.

P and T are directly proportional to each other that if we have two cases we get: P1T2 = P2T1.

In our problem:

P1 = 2200.0 mmHg and T1 = ??? (is needed to be calculated) .

P2 = 2700.0 mmHg and T2 = 39.0 °C + 273 = 312.0 K.

Herein, there is no necessary to convert P into atm.

Then, the temperature in the morning (T1) = P1T2 / P2 = (2200.0 mmHg x 312.0 K) / (2700.0 mmHg) = 254.22 K = -18.77 °C.

6 0

3 years ago

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